DCP 1244 Discrete Mathematics Lecture 20: Planar Graphs

DCP 1244 Discrete Mathematics Lecture 20: Planar Graphs

Tsung Tai Yeh

2021

Outline

I What is a Planar Graph ? I Euler Planar Formula I Graph Coloring

What is a Planar Graph

I A planar graphis an undirected grpah that can be drawn on a plane without any edges crossing.

I Such drawing is called a planar representationof the graph.

The graph K4 K4 drawn with no crossings

Examples of Planar Graphs

I The planar graph of the cube graph

The graph Q3 A planar

representation of Q3

Examples of Planar Graphs

I K1,n and K2,n are planar graphs for all n.

K1, 5 K2, 4

Test Yourself

I Is the graph K3,3 planar ?

- v1 and v2 must connect to v4 and v5. These four edges form a region R₂.

- The edge across v3, v4, v5, splits region into R21 and R22

- We cannot place v6 in any regions to draw a edge without a crossing.

- K3,3 is not a planar graph.

v1 v2 v3

v4 v5 v6

v1 v5

v4 v2

v1 v5

v1 v5

R2 R1 v3 R21

R22 R1

Euler’s Planar Formula

I A planar representation of a graph splits the plane into regions - One of them has infinite area called unbounded region.

I

R1 R2 R3 R4

R1

R2 R3

R4

R5 R6

4 regions

(R4 is unbounded region) 6 regions (R6 is unbounded region)

Euler’s Planar Formula

I Let G be a connected planar graph.

- V = the number of vertices.

- E = the number of edges.

- R = the number of regions.

I Theorem: R = E - V + 2

V = 4, E = 6, R = 4 V = 8, E = 12, R = 6

Euler’s Planar Formula

I Proof Idea:

- Adding edges one by one and keep the subgraph be connected in each step.

- Using induction to show the formula is always satisfied for each subgraph.

I When adding a new edge, it either meet:

- Case 1: two existing vertices.

- Case 2: one existing + one new vertex.

Euler’s Planar Formula

I

R = E – V + 2

E = 1, V = 2, R = 1 Case 2 Case 1

Case 2 Case 1

Case 1

Euler’s Planar Formula

I The degree of a region

- The number of edges on the boundary of a region.

- When an edge occurs twice on the boundary, it contributes two to the degree. e.g. {g , f } and {b, c} in the following graph.

R1

R3 R2

c

b

a d

e g

Test Yourself

I A connected planar simple graph has 20 vertices, each of degree 3. How many regions does a representation of this planar graph split the plane ?

I Solution:

- The sum of the degrees of the vertices is equal to twice the number of edges.

- The sum of the degrees of the vertices: 3v = 3 · 20 = 60.

- 2e = 60, e = 30

- The number of region: R = E − V + 2 = 30 − 20 + 2 = 12.

Euler’s Planar Formula

I Let G be a connected planar simple graph - V is the number of vertices.

- E is the number of edges. 3R I Corollary: E ≤ 3V − 6, where V ≥ 3

I Corollary: G has a vertex of degree not exceeding five.

I Proof:

- Each region has degree greater than or equal to three:

- 2e ≥ 3r - (2/3)e ≥ r - r = e − v + 2

Euler’s Planar Formula

I Prove that K5 are non-planar.

I Proof:

- For K₅, V = 5 and E = 10

- E > 3V − 6, Thus, K₅ is non-planar.

- For K3,3, V = 6 and E = 9

I Corollary: A connected planar simple graph has e edges and v vertices, where v ≥ 3 and no circuits of length three.

e ≤ 2v − 4

I Prove K_3,3 is non-planar

- K3,3 has no circuits of length three.

- e = 9, 2v − 4 = 8, Thus, K3,3 is non-planar.

Kuratowski’s Theorem

I A subdivisionoperation on an edge {u, v } is to create a new vertex w, and replace the edge by two new edges {u, w } and {w , v }.

I

u v

Subdivision on {u, v}

u w v

Kuratowski’s Theorem

I The graph G1 and G2 arehomeomorphic if both can be obtained from the same graph by a sequence of subdivision operations.

I The following graphs are all homeomorphic

G1 G2 G3

Kuratowski’s Theorem

I Show that graph G1 and G2 are homeomorphic

- To obtain G2 from G1, we can use subdivision operations - Case 1: Remove the edge {a, c}, add the vertex f and add the edges {a, f }, {f , c}

- Case 2: Remove the edge {b, c}, add the vertex g, and add the edges {b, g }, {g , c}

- Case 3: Remove the edge {b, g }, add the vertex h, and add the edges {g , h}, {b, h}

a b a b

f h

G1 G2

Kuratowski’s Theorem

I Theorem: A graph G is non-planar if and only if G has a subgraph homeomorphic to K_3,3 or K₅

I Show the graph G is non-planar.

- H is a subgraph of G by deleting h, j, k and all edges incident with these vertices.

- H is homeomorphic to K5 through the subdivision by adding the vertices d, e, and f.

- G is non-planar.

a b

c d f e i h

j

k

a b

c d f e i

a

i c

b

Kuratowski’s Theorem

I Is Petersen graph G planar ?

- H is the subgraph of G by deleting b and three edges that have b as an endpoint.

- H is homeomorphic to K_3,3 through the subdivision operations by deleting {d , h}, add {c, h} and {c, d }, deleting {e, f }, adding {a, e} and {a, f }, deleting {i , j} and adding {g , i } and {g , j}

- The Petersen graph G is not planar.

a

b

e j f g

a

d c f

g

j f d j

Graph Coloring

I A coloringof a simple graph is the assignment of a color to each vertex in a graph

- No two adjacent vertices are assigned the same color.

I The chromatic numberof a graph is the least number of colors needed for a coloring in a graph, denoted by χ(G )

Dual Graph

I To construct the dual graphis based on:

- Each region of the map is represented by a vertex.

- Edges connect two vertices if the regions represented by these vertices have a common border.

A

B

C D

E A

B

D

C E

A Map M Dual Graph of M

The Four Color Theorem

I Theorem: The chromatic number of a planar graph is no greater than four.

I What are the chromatic numbers of the graph G ? - The chromatic number of G is at least 3, because the vertex a, b, c must be assigned different colors.

a

b

c d

e

f

g a

b

c d

e

f

g

Chromatic Number

I What is the chromatic number of the graph K_n ? - K_n is not planar when n ≥ 5. That is χ(K_n) = n.

I What is the chromatic number of graph Km,n? - Km,n is a bipartite graph. Thus, χ(Km,n) = 2.

a

c

d e

b a b

c d e

A coloring of K2,3